Acceleration changing over time

[nicklasaau]

Hey, i am having trouble doing the calculation on a dynamic system.

My acceleration is not constant during, and because of that my speed of the particle is changing from the original, and i want to know how long my particle has moved in a short periode of time Δt and what the velocity is. with the math i have learned so far, i end up with only taking into account the end velocity or the beginning velocity, instead of calculating it is changing


Know is
Beginning condition: (distance, speed)
Travel: (acceleration at a specific time, that changes)
I want to know end distance and end velocity.

At the end i need to rewrite it to a transfer function, so i have to do linearization if that can help in the calculation :)

 

[mikeph]

A general solution will require calculus, but it's possible you won't need this.

velocity: v(t) = ∫a(t)dt + v0
displacement: x(t) = ∫(v(t))dt + x0 = ∫(∫a(t)dt + v0)dt + x0

v0, x0 are the velocity and displacement at t=0 (initial conditions).

You will need to integrate the acceleration a(t) twice to get a distance, and the method of solution will depend on the form of a(t). Do you know anything about the form of a(t)?

 

[nicklasaau]

The object is a levitating beam, and because of that the acceleration is a function of time and distance



a(t,x) = (μ * N^2 * Ag)/4 * [I*sin(ωt + ρ)]^2 / x^2, or something like that

 

[Chestermiller]

A general solution will require calculus, but it's possible you won't need this.

velocity: v(t) = ∫a(t)dt + v0
displacement: x(t) = ∫(v(t))dt + x0 = ∫(∫a(t)dt + v0)dt + x0

v0, x0 are the velocity and displacement at t=0 (initial conditions).

You will need to integrate the acceleration a(t) twice to get a distance, and the method of solution will depend on the form of a(t). Do you know anything about the form of a(t)?

That double integral involving a(t) can be reexpressed as a single integral by integrating by parts to obtain:

\int_0^t(t-\lambda)a(\lambda) d\lambda

where λ is a dummy variable of integration.